1. **Problem Statement:** We are given that 60% of pupils have internet access at home. From a group of 8 students chosen at random, we want to find: a. The probability exactly 5 have internet access. b. The probability at least 6 have internet access. 2. **Formula and Explanation:** This is a binomial probability problem where: - Number of trials $n = 8$ - Probability of success (internet access) $p = 0.6$ - Probability of failure $q = 1 - p = 0.4$ The binomial probability formula is: $$P(X = k) = \binom{n}{k} p^k q^{n-k}$$ where $k$ is the number of successes. 3. **Calculations:** **a. Exactly 5 have access:** $$P(X=5) = \binom{8}{5} (0.6)^5 (0.4)^3$$ Calculate the combination: $$\binom{8}{5} = \frac{8!}{5!3!} = \frac{40320}{120 \times 6} = 56$$ Calculate powers: $$(0.6)^5 = 0.07776$$ $$(0.4)^3 = 0.064$$ Multiply all: $$P(X=5) = 56 \times 0.07776 \times 0.064 = 56 \times 0.00497664 = 0.2787$$ **b. At least 6 have access:** This means $P(X \geq 6) = P(X=6) + P(X=7) + P(X=8)$ Calculate each term: - For $X=6$: $$P(X=6) = \binom{8}{6} (0.6)^6 (0.4)^2$$ $$\binom{8}{6} = 28$$ $$(0.6)^6 = 0.046656$$ $$(0.4)^2 = 0.16$$ $$P(X=6) = 28 \times 0.046656 \times 0.16 = 28 \times 0.007465 = 0.209$$ - For $X=7$: $$P(X=7) = \binom{8}{7} (0.6)^7 (0.4)^1$$ $$\binom{8}{7} = 8$$ $$(0.6)^7 = 0.0279936$$ $$(0.4)^1 = 0.4$$ $$P(X=7) = 8 \times 0.0279936 \times 0.4 = 8 \times 0.01119744 = 0.08958$$ - For $X=8$: $$P(X=8) = \binom{8}{8} (0.6)^8 (0.4)^0$$ $$\binom{8}{8} = 1$$ $$(0.6)^8 = 0.016777216$$ $$(0.4)^0 = 1$$ $$P(X=8) = 1 \times 0.016777216 \times 1 = 0.01678$$ Sum these probabilities: $$P(X \geq 6) = 0.209 + 0.08958 + 0.01678 = 0.31536$$ 4. **Final answers:** - Probability exactly 5 have access: **0.279** (rounded to 3 decimals) - Probability at least 6 have access: **0.315** (rounded to 3 decimals)